As a point of curiosity, I found myself wondering when a complex number to an integer power creates a real number. For the sake of completeness, I also looked at when the result is a fully imaginary number.
More rigorously, define \(\mathbb{C}^*\) as the set of complex numbers \(a + bi\) where both \(a\) and \(b\) are non-zero, and \(k\in\mathbb{N}^*\). For what values \(x \in \mathbb{C}^*\) is \(x^k \in \mathbb{R} \lor \mathbb{I}\)?
k = 1
This is the trivial case. Since \(x^1 = x\), there are no real or imaginary cases.
k = 2
\((a + bi)^2 = a^2 + 2abi + b^2i^2 = a^2 + 2abi – b^2 = (a^2 – b^2) + 2abi\).
This is real when the coefficient of \(i\), i.e., \(2ab\) is equal to 0. This is only true when \(a = 0\) or \(b = 0\), so it is not true for any element of \(\mathbb{C}^*\).
This is imaginary when \(a^2 – b^2 = 0 \Rightarrow a^2 = b^2\). That is, \((a \pm ai)^2 \in \mathbb{C}^*\).
k = 3
\((a + bi)^3 = a^2 + 3a^2bi – 3ab^2 – b^3i = a(a^2 – 3b^2) + b(3a^2 – b^2)i\).
This is real when \(b(3a^2 – b^2) = 0\). If \(b = 0, x\notin\mathbb{C}^*\), so we ignore that case. \(3a^2 – b^2 = 0 \Rightarrow 3a^2 = b^2\), so \(b = \pm\sqrt{3} a\). Hence, if \(x\in\mathbb{C}^*\), \(x = a \pm a\sqrt{3}i\).
Similarly, this is imaginary when \(a(a^2 – 3b^2) = 0\), so \(b = \pm\frac{\sqrt{3}}{3} a\) and \(x = a \pm\frac{a\sqrt{3}}{3}i\).
k = 4
Using similar methods, we get: \((a + bi)^4 = (a^4 – 6a^2b^2 + b^4) + 4ab(a^2 – b^2)i\).
This is real when \(a^2 = b^2\), so \(x = a \pm ai\).
Replacing \(a^2 = c\) and \(b^2 = d\) gives us the real part of \(d^2 – 6dc + c^2\). Applying the quadratic formula with \(a = 1\), \(b = -6c\), and \(d = c^2\) yields \(c = \frac{6c \pm \sqrt{36c^2 – 4c^2}}{2} = 3c \pm c\sqrt{8} = (3 \pm 2\sqrt{2}) c\). Note that since the quadratic is symmetrical, that means that \(3 + 2\sqrt{2} = \frac{1}{3 – 2\sqrt{2}}\).
Since \(a^2 = c\) and \(b^2 = d = (3 \pm 2\sqrt{2})c \), \(a = \pm\sqrt{c}\) and \(b = \pm(3 \pm 2\sqrt{2})\sqrt{c} = \pm(3 \pm 2\sqrt{2})a\).
This is imaginary, then, when \(x = a \pm(3 \pm 2\sqrt{2}) ai\).
k = 5
This time, we get: \((a + bi)^5 = a(a^4 – 10a^2b^2 + 5b^4) + b(5a^4 – 10a^2b^2 + b^4)i\)
This is real when \(5a^4 – 10a^2b^2 + b^4 = 5c^2 – 10cd + d^2 = 0\).
\(d = \frac{10c \pm\sqrt{100c^2 – 20c^2}}{10} = \frac{10c \pm 4c\sqrt{5}}{2} = 5 \pm 2\sqrt{5} c\), so \(b = \pm\sqrt{5 \pm 2\sqrt{5}}a\).
This is real, then, when \(x = a \pm \sqrt{5 \pm 2\sqrt{5}}ai\).
This is imaginary when \(a^4 – 10a^2b^2 + 5b^4 = c^2 – 10cd + 5d^2 = 0\).
\(d = \frac{10c \pm \sqrt{100c^2 – 20c^2}}{10} = \frac{10c \pm 4c\sqrt{5}}{10} = (1 \pm \frac{2\sqrt{5}}{5})c\), so \(b = \pm\sqrt{1\pm\frac{2\sqrt{5}}{5}}a\).
This is imaginary, then, when \(x = a \pm\sqrt{1\pm\frac{2\sqrt{5}}{5}}ai\).
k = 6
This time, we get: \((a + bi)^6 = (a^6 – 15a^4b^2 + 15a^2b^4 – b^6) + 2ab(3a^4 – 10a^2b^2 + 3b^4)i\)
This is real when \(3a^4 – 10a^2b^2 + 3b^4 = 3c^2 – 10cd + 3d^2 = 0\).
\(d = \frac{10c \pm \sqrt{100c^2 – 36c^2}}{6} = \frac{10c \pm 8c}{6} = \frac{5\pm 4}{3}c\), so \(b = \pm\sqrt{\frac{5\pm 4}{3}}a\).
This is real, then, when \(x = a \pm\sqrt{\frac{5\pm 4}{3}}ai\).
This is imaginary when \(a^6 – 15a^4b^2 + 15a^2b^4 – b^6 = c^3 – 15c^2d + 15cd^2 – d^3 = 0\).
We can factor \(c^3 – 15c^2d + 15cd^2 – d^3 = (c – d)(c^2 – 14cd + d^2)\). \(c – d = 0\) when \(c = d\), so \(b = \pm a\) is one possibility. Let’s look at the other case.
\(c^2 – 14cd + d^2 = 0\) when \(d = \frac{14c \pm\sqrt{196c^2 – 4c^2}}{2} = 7c \pm\sqrt{48}c = (7 \pm 4\sqrt{3})c\), so \(b = \pm\sqrt{7 \pm 4\sqrt{3}}a\).
This is imaginary, then, when \(x \in \{a \pm ai, a \pm\sqrt{7\pm 4\sqrt{3}}ai\} \).
Summary
At \(k = 7\), we have the cubics \(c^3 – 21c^2d + 35cd^2 – 7d^3\) and \(7c^3 – 35c^2d + 21cd^2 – d^3\), which do not have factors, so this is a good place to stop.
Factoring out \(a\), here are the cases where \(x\in\mathbb{C}^*\) yields \(x^k \in \mathbb{R} \lor \mathbb{I}\):
| k | \(x^k\in\mathbb{R}\) | \(x^k\in\mathbb{I}\) |
| 1 | (None) | (None) |
| 2 | (None) | \(1 \pm i\) |
| 3 | \(1\pm\sqrt{3}i\) | \(1\pm\frac{\sqrt{3}}{3}i\) |
| 4 | \(1\pm i\) | \(1 \pm(3 \pm 2\sqrt{2})i\) |
| 5 | \(1\pm\sqrt{5\pm 2\sqrt{5}}i\) | \(1\pm\sqrt{1\pm\frac{2\sqrt{5}}{5}}i\) |
| 6 | \(1\pm\sqrt{\frac{5\pm 4}{3}}i\) | \(\{1\pm i, 1\pm\sqrt{7\pm 4\sqrt{3}}i\}\) |